Solving Separating Variables Differential Equations

[KevinL]

I would just like to have my first two questions checked, and on the last one I'm not even quite sure how to start it.

Find the general solution of the differential equation.

1) dy/dt = t^4*y

dy/y = t^4 dt

Integrate both sides:

ln|y| = (t^5)/5 +c

Raise both sides by e to get rid of ln:

y = +/- ce^(t^5/5)

Is y = 0 also a solution (equilibrium)?

2) dy/dt = 2-y

dy/(2-y) = 1 dt

integrate both sides:

-ln|2-y| = t+c
ln|2-y| = -t+c

Raise both sides by e to get rid of ln:

|2-y| = ce^-t
y=ce^-t +2

Is y=2 also a solution (equilibrium)?

3) dy/dt = 1/(ty + t + y + 1)

Im not even sure how to get this in the form dy/dt = f(t)g(y)

 

[rock.freak667]

I would just like to have my first two questions checked, and on the last one I'm not even quite sure how to start it.

Find the general solution of the differential equation.

1) dy/dt = t^4*y

dy/y = t^4 dt

Integrate both sides:

ln|y| = (t^5)/5 +c

Raise both sides by e to get rid of ln:

y = +/- ce^(t^5/5)

Is y = 0 also a solution (equilibrium)?

This is good.

so if y=0 is a solution, then 0=ce^t5/5

is there any value of t that satisfies this?

2) dy/dt = 2-y

dy/(2-y) = 1 dt

integrate both sides:

-ln|2-y| = t+c
ln|2-y| = -t+c

Raise both sides by e to get rid of ln:

|2-y| = ce^-t
y=ce^-t +2

Is y=2 also a solution (equilibrium)?

This is good as well. Similar, with the one above, 2=ce^-t+2. Is there any t to satisfy this?

EDIT: even though it is correct, you should use a different letter for the constant c, just so it isn't technically wrong.

3) dy/dt = 1/(ty + t + y + 1)

ty+t+y+t ≡ ty+y +t+1 ≡ (t+1)y+(t+1)

see it now?

 

[KevinL]

so if y=0 is a solution, then 0=ce^t5/5

is there any value of t that satisfies this?

t=0. The way I thought of it originally was that we had the integral dy/y, so if y=0 you obviously would get problems. Is this a fair way to think of it?

Similar, with the one above, 2=ce^-t+2. Is there any t to satisfy this?

t=0. Similarly, when you have dy/(2-y), y=2 would give you problems. Again, is this a fair way to find the solution y=2?

ty+t+y+t ≡ ty+y +t+1 ≡ (t+1)y+(t+1)

see it now?

Ah, of course. So...

(y+1) dy = 1/(t=1) dt

Integrate both sides:

y^2/2 + y = ln|t+1|

e^(y^2/2)*e^y -1 = t

Now we're stuck (or at least I am stuck! . The HW instructions mentioned that this could happen, but to get as "far as you can". Does this suffice for an answer?

 

[rock.freak667]

Say we had e^x=2 and we wanted to find x. We'd take ln of both sides

lne^x=ln2 => x*lne=ln2 =>x=ln2


so now for e^x=0, take ln lne^x=ln(0). We have a problem now, ln(0) does not exist. So there is no x that satisfies e^x=0. Better now?

 

[KevinL]

Makes sense now. Thank you for the help :)